Ex. 1. How many minutes does Aditya take to cover a distance of 400 m, if he runs at a speed of 20 km/hr?
Sol. Aditya’s speed = 20 km/hr = {20 * 5} m/sec = 50 m/sec
18 9
\Time taken to cover 400 m= { 400 * 9 } sec =72 sec = 1 12 min 1 1 min.
50 60 5
Ex. 2. A cyclist covers a distnce of 750 m in 2 min 30 sec. What is the speed in km/hr of the cyclist?
Sol. Speed = { 750 } m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr
150 5
Ex. 3. A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.
Sol. Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.
Then , 3x = 4y => x = 4 y => 4x = 16 y.
3 3
\ Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time
= 4x : 5y = 16 y : 5y =16 : 5 = 16:15
3 3
Ex. 4.While covering a distance of 24 km, a man noticed that after walking for 1 hour and 40 minutes, the distance covered by him was 5 of the remaining distance. What was his speed in metres per second?
7
Sol. Let the speed be x km/hr.
Then, distance covered in 1 hr. 40 min. i.e., 1 2 hrs = 5x km
3 3
Remaining distance = { 24 – 5x } km.
3
\ 5x = 5 { 24 - 5x } ó 5x = 5 { 72-5x } ó 7x =72 –5x
3 7 3 3 7 3
ó 12x = 72 ó x=6
Hence speed = 6 km/hr ={ 6 * 5 } m/sec = 5 m/sec = 1 2
18 3 3
Ex. 5.Peter can cover a certain distance in 1 hr. 24 min. by covering two-third of the distance at 4 kmph and the rest at 5 kmph. Find the total distance.
Sol. Let the total distance be x km . Then,
2 x 1 x
3 + 3 = 7 ó x + x = 7 ó 7x = 42 ó x = 6
4 5 5 6 15 5
Ex. 6.A man traveled from the village to the post-office at the rate of 25 kmph and walked back at the rate of 4 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.
Sol. Average speed = { 2xy } km/hr ={ 2*25*4 } km/hr = 200 km/hr
x+y 25+4 29
Distance traveled in 5 hours 48 minutes i.e., 5 4 hrs. = { 200 * 29 } km = 40 km
5 29 5
Distance of the post-office from the village ={ 40 } = 20 km
2
Ex. 7.An aeroplane files along the four sides of a square at the speeds of 200,400,600 and 800km/hr.Find the average speed of the plane around the field.
Sol. :
Let each side of the square be x km and let the average speed of the plane around the field by y km per hour then ,
x/200+x/400+x/600+x/800=4x/yó25x/2500ó4x/yóy=(2400*4/25)=384
hence average speed =384 km/hr
Ex. 8.Walking at 5 of its usual speed, a train is 10 minutes too late. Find its usual time to cover the journey.
7
Sol. :New speed =5/6 of the usual speed
New time taken=6/5 of the usual time
So,( 6/5 of the usual time )-( usual time)=10 minutes.
=>1/5 of the usual time=10 minutes.
ð usual time=10 minutes
Ex. 9.If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
Sol. Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5-x/6=1/5<=>6x-5x=6
óx=6
Hence, the required distance is 6 km
Ex. 10. A and B are two stations 390 km apart. A train starts from A at 10 a.m. and travels towards B at 65 kmph. Another train starts from B at 11 a.m. and travels towards A at 35 kmph. At what time do they meet?
Sol. Suppose they meet x hours after 10 a.m. Then,
(Distance moved by first in x hrs) + [Distance moved by second in (x-1) hrs]=390.
65x + 35(x-1) = 390 => 100x = 425 => x = 17/4
So, they meet 4 hrs.15 min. after 10 a.m i.e., at 2.15 p.m.
Ex. 11. A goods train leaves a station at a certain time and at a fixed speed. After ^hours, an express train leaves the same station and moves in the same direction at a uniform speed of 90 kmph. This train catches up the goods train in 4 hours. Find the speed of the goods train.
Sol. Let the speed of the goods train be x kmph.
Distance covered by goods train in 10 hours= Distance covered by express train in 4 hours
10x = 4 x 90 or x =36.
So, speed of goods train = 36kmph.
Ex. 12. A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?
Sol. Relative speed of the policeman = (10-8) km/hr =2 km/hr.
Time taken by police man to cover 100m 100 x 1 hr = 1 hr.
1000 2 20
In 1 hrs, the thief covers a distance of 8 x 1 km = 2 km = 400 m
20 20 5
Ex.13. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?
Sol. Let the distance be x km. Then,
( Time taken to walk x km) + (time taken to ride x km) =37 min.
( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
But, the time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (74-55)min =19 min.